# Finding the Minimum Work Done for a Thermodynamic Process

1. Nov 14, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
Given two indistinguishable objects at the same initial temperature $T_i$, calculate the minimum work done by a refrigerator functioning between the two objects till one of the objects reaches a new temperature $T_2$, assume constant heat capacities and constant pressure.

a) Prove $W^{min}_{on} = C_p[(\frac{T_i^2}{T_2})+T_2-2T_i]$

b) Now prove it using $W_{by} \leq -\Delta A$, are there any other requirements necessary for the results from a) and b) to agree?

2. Relevant equations
$dU = \delta Q_{in} + \delta W_{on}$
$\delta W_{quas} = -PdV$
$dS=\frac{\delta Q_{quas}}{T}$
$C_P=\left(\frac{\delta Q_{rev}}{dT}\right)_P$
3. The attempt at a solution

For part a:

Since it's constant pressure $\delta Q_{rev} = C_pdT$, now since there's two objects I think the total heat is $Q_{revT} = C_p\int_{T_i}^{T_2}dT + C_p\int_{T_i}^{T_h}dT$ where $T_h$ is the final temperature of the object that isn't being cooled, I believe the heat of this object goes up. Factoring and integrating gives $Q_{revT} = C_p(T_2-T_i+T_h-T_i) = C_p(T_2-2T_i+T_h)$

Edit: Ok I've managed to get the answer but frankly I don't know how to justify what I've done.

For the change in entropy we have $$\Delta S = C_p\int_{T_i}^{T_2}\frac{dT}{T}+C_p\int_{T_i}^{T_h}\frac{dT}{T} = C_p\ln(\frac{T_2}{T_i})+C_p\ln(\frac{T_h}{T_i})=C_p\ln(\frac{T_2T_h}{T_i^2})$$

Now if I set the change in entropy equal to zero I can get $T_h = \frac{T_i^2}{T_2}$ which plugging into the above expression gives the answer but I don't see why $\delta S = 0$ and it seems volume remains constant as well which gives the result $W_{on} = -Q_{in}$

Last edited: Nov 14, 2016
2. Nov 14, 2016

### Staff: Mentor

What part of it are you uncomfortable with? You are aware that, when you are referring to the amount of work, you are talking about the work that the working fluid in the refrigeration cycle has to do, not to work done by the two objects, correct?

3. Nov 14, 2016

### Potatochip911

Oh I actually was not aware of that. So the change in entropy is zero because the heat leaving the hot object is equal in magnitude for the heat entering the cold cold object ($dS= \frac{dQ}{T})$)? The only thing then that concerns me with that logic is I believe T refers to the heat of the object so the hot one has a larger denominator throughout the integral.

4. Nov 14, 2016

### Staff: Mentor

No. Less heat leaves the cold object than enters the hot object. That's why the two entropy changes are the same.
$$Q_h=C(T_h-T_i)$$
$$Q_C=C(T_2-T_i)$$
The sum of these two heats is equal to the work that the working fluid has to do.

5. Nov 14, 2016

### Potatochip911

Is this a consequence of work being put into the refrigerator?

How do we know that the internal energy of the working fluid doesn't change?

Edit: Ok I see now that we didn't set internal energy equal to zero for the working fluid but the work done by the fluid is 0 and then the sum of the internal energies of the working fluid is equal to the work. I don't understand why that sum gives work

Last edited: Nov 14, 2016
6. Nov 14, 2016

### Staff: Mentor

A refrigerator works by pumping heat from a cold object and rejecting a greater amount of heat to a hot object (usually, the surrounding room).

The working fluid is subjected to a cycle. Actually, as the hot body gets hotter and the cold body gets colder, the size of the cycle applied to the working fluid expands gradually, but the deviation from a perfect cycle in any pass is regarded as negligible. So, as in any cycle, since internal energy is a state function, it does not change over a cycle.
Since the change in internal energy over any cycle is zero, the work done on the working fluid is equal to the heat discharged to the hot reservoir minus the heat removed from the cold reservoir.

7. Nov 14, 2016

### Potatochip911

Thanks this all makes a lot of sense now!